Chemistry 10 2014-2015

MOLECULAR FORMULAS
 If the empirical formula is different from the molecular formula, the molecular formula will always be a simple multiple of the empirical formula.
 EX: The empirical formula for hydrogen peroxide is HO; the molecular formula is
H
2
O
2
.
 In both formulas, the ratio of oxygen to hydrogen is 1:1

DETERMINE THE EMPIRICAL FORMULAS FOR
EACH OF THE FOLLOWING MOLECULAR
FORMULAS.
1. C
6
H
18
…..______________
2. H
2
O
2
……______________
3. Hg
2
Cl
2
…..______________
4. C
3
H
6
…….______________
5. Na
2
C
2
O
4
..._____________
6. H
2
O......________________
7. C
4
H
8
….._______________
8. C
4
H
6
…..________________
9. C
7
H
12
…._______________
10. CH
3
COOH….___________

REVIEW: CALCULATING
EMPIRICAL FORMULAS
 Use the following poem to remember the steps:
Percent to mass
Mass to moles
Divide by small
Multiply ‘til whole

EXAMPLE PROBLEM
 Methyl acetate is a solvent commonly used in some paints, inks, and adhesives.
Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen.

STEP ONE: PERCENT TO MASS
 Let’s assume we have a 100. g sample of methyl acetate. This means that each element’s percent is also the number of grams of that element.
 48.64% C = 48.64 g C
 8.16% H = 8.16 g H
 43.20% O = 43.20 g O

STEP TWO: MASS TO MOLES
 Convert each mass into moles using the molar mass of each element.
 48.64 g C x 1 mol C = 4.050 mol C
12.011 g
 8.16 g H x 1 mol H = 8.10 mol H
1.008 g
 43.20 g O x 1 mol O = 2.700 mol O
15.999 g

STEP THREE: DIVIDE BY SMALL
 Oxygen accounts for the smallest number of moles in the formula, so divide each element by oxygen’s number of moles: 2.700 mol
 Carbon: 4.050 mol / 2.700 mol = 1.500 = 1.5
 Hydrogen: 8.10 mol / 2.700 mol = 3.00 = 3
 Oxygen: 2.700 mol / 2.700 mol = 1.000 = 1
 Remember, we will want whole-number ratios

STEP FOUR: MULTIPLY ‘TIL WHOLE
 In the previous slide, the ratio of C:H:O is 1.5:3:1
 We need a whole-number ratio, so we can
multiply everything by 2 to get rid of the 1.5
 C  3, H  6, O  2
So, the final empirical formula is C
3
H
6
O
2

CALCULATING MOLECULAR
FORMULAS
 Step 1 – Calculate the empirical formula (if needed)
 Step 2 – GIVEN molecular mass (experimental) calculate the multiplier
𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑚𝑎𝑠𝑠
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠
 Step 3 – Multiply the empirical formula subscripts by the multiplier found in Step 2

EXAMPLE PROBLEM
 Succinic acid is a substance produced by lichens.
Chemical analysis indicates it is composed of
40.68% carbon, 5.08% hydrogen, and 52.24% oxygen and has a molar mass of 118.1 g/mol.
Determine the empirical and molecular formulas for succinic acid.

STEP ONE: FIND EMPIRICAL
FORMULA
 40.68 % C = 40.68 g C x 1mol C = 3.387 mol C
12.011 g
 5.08 % H = 5.08 g H x 1 mol H = 5.04 mol H
1.008 g
 54.24% O = 54.24 g O x 1 mol O = 3.390 mol O
15.999 g
C: H:O
3.387 mol : 5.04 mol : 3.390 mol
3.387 mol 3.387 mol 3.387 mol
1 : 1.48: 1  1 : 1.5 : 1, multiply by 2  C
2
H
3
O
2

STEP TWO: DIVIDE MOLAR
MASSES
 Molar mass empirical formula
(2 x 12.0 g/mol) + (3 x 1.01 g/mol) + (2 x 16.0 g/mol)
= 59.0 g/mol
 Given (experimental) molar mass
= 118.1 g/mol
 Multiplier = 118.1 g/mol = 2
59.0 g/mol

STEP THREE: USE MULTIPLIER
 Empirical Formula = C
2
H
3
O
2
 x 2 from step two
 Molecular formula = C
4
H
6
O
4

PRACTICE QUESTION 2
What is the empirical and molecular formula of Vitamin D
3 if it contains
84.31% C, 11.53% H, and 4.16% O, with a molar mass of 384.0 g/mol?

THINKING QUESTIONS
1) What information must a chemist obtain in order to determine the empirical formula of an unknown compound?
2) What information must a chemist have to determine the molecular formula of a compound?